Yes, it's 6(N-1) (for N>1; 1 for N=1 is of course trivial). Essentially the same arguments apply in each case; the simpler argument with the center lines for the cube, and the plane-counting argument for the torus (which of course implies the cube result). And essentially the same solution, packing the bricks into 6 of the corners with different stacks oriented differently, achieves that maximum. I think this can actually be generalized into packing NxNx1 bricks into an MxMxM cube (or torus), with 1 < M/2 < N <= 3M/4, the maximum being 6(M-N), but I'm not 100% sure at the moment. (When N > 3M/4, you do better to put a single stack all the way through the cube, and fit in two small stacks on the sides.) Can anybody find a brick-packing problem where you can fit more bricks into the (3D) torus than you can into the cube? Bricks are AxBxC, not arbitrary polycubes. How about the 2D version? Franklin T. Adams-Watters -----Original Message----- From: dasimov@earthlink.net Do you also know the story for how many NxNx1's can fit in a (2N-1)-cube, as you suggested? (Or torus?) ___________________________________________________ Try the New Netscape Mail Today! Virtually Spam-Free | More Storage | Import Your Contact List http://mail.netscape.com